Determining the minimal polynomial over $\Bbb{Q}$ I was working on a homework assignment from Hungerford:

Find the minimal polynomial of the element $\sqrt{1+\sqrt{5}}$ over $\Bbb{Q}$.

Naturally the solution would be the polynomial with roots 
$$ \pm \sqrt{1 \pm \sqrt{5}} $$ 
Which is found as 
$$ x = \pm \sqrt{1 \pm \sqrt{5}}  \rightarrow x^2 -1 = \pm \sqrt{5} \rightarrow (x^2-1)^2 = 5 \rightarrow $$
$$\text{Minimal Polynomial = } (x^2-1)^2-5$$
Problem is, I quite frankly don't know how to prove this. Hypothetically what if, in the same vein that 
$$ \sqrt{5 + 2\sqrt{6}} = \sqrt{2} + \sqrt{3}$$ 
There exists some subtle factorization for $\sqrt{1 + \sqrt{5}}$ with the additionaly property that it is the root of a cubic or quadratic. How do I definitely rule out any of those cases?
 A: Maybe there's a nicer way, but you could consider the intermediate extension: 
$$\Bbb Q \subseteq \Bbb Q(1+\sqrt 5)\subseteq \Bbb Q(\sqrt{1+\sqrt 5})$$
$1+\sqrt 5$ has degree $2$ over $\Bbb Q$ because it's not rational and it's a root of $x^2-2x-6 \in \Bbb Q[x]$.
Because of the multiplicative property, $[\Bbb Q(\sqrt{1+\sqrt 5}): \Bbb Q]$ must be divisible by $2$,  and it is less than $4$ because you showed $\sqrt{1+\sqrt 5}$ is a root of a degree-four polynomial. Hence, it cannot be $3$.
Now you only have to show that $\sqrt{1+\sqrt 5}$ is not an element of $\Bbb Q(1+\sqrt 5)$, and you can show this by contradiction.
Since $\Bbb Q(1+\sqrt 5)$ has degree $2$ over $\Bbb Q$, it has $1, 1+\sqrt 5$ as a basis, so you can try to write 
$$\sqrt{1+\sqrt 5} = a+b(1+\sqrt 5) \text{ for some } a,b \in \Bbb Q$$
and derive a contradiction (which might be somewhat tedious).
A: It is enough to show that $(x^2-1)^2-5$, that is, $x^4-2x^2-4$ is irreducible over the rationals. There are no rational roots, so we only need to rule out factorization as a product of quadratics with integer coefficients. 
Since there is no $x^3$ term. we can confine attention to factorizations $(x^2+ax+b)(x^2-ax+c)$. If $a=0$, we get $b+c=-2$, $bc=-4$, impossible in integers. If $a\ne 0$, then because there is no $x$ term we need $b=c$, but then $bc$ cannot be $-4$.   
